The bob of a simple pendulum (mass $m$ and length $l$) is dropped from a horizontal position. It strikes a block of the same mass $m$ placed on a horizontal frictionless table. The collision is elastic. The kinetic energy of the block after the collision will be

  • A
    $2\, mgl$
  • B
    $mgl/2$
  • C
    $mgl$
  • D
    $0$

Explore More

Similar Questions

$A$ ball of mass $m$ is thrown at a wall with speed $v$ at an angle $\theta$ with the normal to the wall. If the coefficient of restitution is $e$,what will be the magnitude and direction of the velocity of the ball after the collision with respect to the wall?

Difficult
View Solution

$A$ body of mass $m$ moving with velocity $v$ collides with a stationary body of mass $2m$. The fraction of kinetic energy lost by the body of mass $m$ is:

Difficult
View Solution

On a frictionless surface,a block of mass $M$ moving at speed $V$ collides elastically with another block of same mass $M$ which is initially at rest. After collision,the first block moves at an angle $\theta$ to its initial direction and has a speed $\frac{V}{3}$. The second block's speed after the collision is:

Which of the following is not a perfectly inelastic collision?

$A$ steel sphere of radius $1.2 \ cm$ collides with a second steel sphere at rest. If the collision is elastic and after the collision the first sphere continues to move in its initial direction with a velocity of $\frac{7}{9}$ times its initial velocity,then the radius of the second sphere is (in $cm$)

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo